8*9-6x=18

(x+3)/(x+2)+(x+9)/(x+8)=(x+5)/(x+4)+(x+7)/(x+6)1+(x+5)/(x+2)(x+8)=1+(x+5)/(x+4)(x+6)x=-5再问：第二步那个1是怎么

将x-7/x-9分解成1+2/(x-9)其他分式同理则原方程等价于1/x-9+1/x-5=1/x-6+1/x-81/x-6-1/x-5=1/x-9-1/x-81/(x-5)(x-6)=1/(x-8)(

设a=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)那么y=a*(x-10);那么y^=a^*(x-10)+a*(x-10)^=a^*(x-10)+a那么y

1）（x-3）/（x-2）-（x-5）/（x-4）=（x-7）/（x-6）-（x-9）/（x-8）化简得【（x-3）（x-4）-（x-5）/（x-2）】/【（x-2）（x-4)】=【（x-7）（x-8

 再问：额、不懂再答： 再答：后面的看做一个整体再问：好的吧、谢谢大神再答：回来的话，请采纳再问：啊、突然明白了呢。。。

[1-1/(x-4)]+[1-1/(x-8)]=[1-1/(x-7)]+[1-1/(x-5)]1/(x-4)+1/(x-8)=1/(x-7)+1/(x-5)(x-4+x-8)/(x-4)(x-8)=(

(x+7/x+6)+(x+9/+x+8)=(x+10/x+9)+(x+6/x+5)(x+9/+x+8)-(x+10/x+9)=(x+6/x+5)-(x+7/x+6)(((x+9)^2-(x+8)(x+

1.(x+7/x+6)+(x+9/+x+8)=(x+10/x+9)+(x+6/x+5)(x+9/+x+8)-(x+10/x+9)=(x+6/x+5)-(x+7/x+6)(((x+9)^2-(x+8)(

x(1+2+...+9)=x(9-8-7-...-1)x=0记得采纳啊

/>因为：x-5/x-6+x-8/x-9=x-7/x-8+x-6/x-7,所以：1+1/(x-6)+1+1/(x-9)=1+1/(x-8)+1+1/(x-7)即：1/(x-6）+1/(x-9)=1/(

(1)(X+7)/(X+6)+(X+9)/(X+8)=(X+10)/(X+9)+(X+6)/(X+5)这类方程重在运算的技巧性,观察分母7+9=10+6;6+8=5+9所以先移项:(X+7)/(X+6

即1/(x-5)+1/(x-9)=1/(x-6)+1/(x-8)(2x-14)[1/(x-5)(x-9)-1/(x-6)(x-8)]=01/(x-5)(x-9)-1/(x-6)(x-8)≠0所以2x-

设x1=4-√3,x2=4+√3,是方程X^2-8X+13=0的两根所以X1^2-8X1+15=2X^4-6X^3-2X^2+18X+23=(X+1)^2*(X^2-8X+13)+10=10所以原式=

这道题不用那么复杂.先移项.(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/( x-9)再通分.（x-4)(x-6)/(x-5)(x-6)-(x-5)(x

左右两边同乘以(X-5)(X-6)(X-8)(X-9)得到(X-4)(X-6)(X-8)(X-9)-(X-5)(X-5)(X-8)(X-9)=(X-5)(X-6)(X-7)(X-9)-(X-5)(X-

(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/(x-9)1+1/(x-5)-1-1/(x-6)=1+1/(x-8)-1-1/(x-9)1/(x-5)-1/(x-6

7/x+8/x-4/x-5/x=-8-9+5+66/x=-6x=-1

x-4/x-5-x-5/x-6=x-7/x-8-x-8/x-91+1/x-5-1-1/x-6=1+1/x-8-1-1/x-91/x-5-1/x-6=1/x-8-1/x-9-1/(x-5)(x-6)=-

(X-4)/(X-5)-(X-5)/(X-6)=(X-7)/(X-8)-(X-8)/(X-9)[(x-4)(x-6)-(x-5)2]/(x-5)(x-6)=[(x-7)(x-9)-(x-8)2]/(x

(x-4)/(x-5)+(x-8)/(x-9)=(x-7)/(x-8)+(x-5)/(x-6)1+1/(x-5)+1+1/(x-9)=1+1/(x-8)+1+1/(x-6)1/(x-5)+1/(x-9